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A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

Short Answer

Expert verified
(D) It will increase because the gas molecules will be more dispersed in the larger flask. As the volume increases while the temperature remains constant, the entropy increases due to greater dispersal of gas molecules.

Step by step solution

01

Understanding the concept of Entropy

Entropy is a measure of the disorder or randomness of a system. In this exercise, we are considering a case where a gas is transferred from a smaller to a larger container, keeping the temperature constant.
02

Impact of Volume on Entropy

If the volume in which the gas is contained increases, there will be more space for the gas particles to move about. This would lead to higher randomness or disorder in the gas distribution, which in turn will lead to an increase in entropy.
03

Identifying the right answer

Based on our analysis in the previous steps, the answer is (D) It will increase because the gas molecules will be more dispersed in the larger flask. This is because as volume increases, at a constant temperature, the gas particles have increased space to move leading to an increase in entropy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Laws
Gas laws help us understand the behavior of gases in different scenarios. They describe how variables like volume, pressure, temperature, and quantity are interrelated in gases. Imagine you're studying how a balloon behaves when you take it outside on a chilly day or how it expands when it's heated.

One important gas law relevant here is Boyle’s Law, which states that the volume of gas is inversely proportional to pressure at a constant temperature. This means if you increase the volume available to a gas, its pressure will decrease, assuming the temperature remains unchanged.

Another crucial principle is the Ideal Gas Law, expressed as \( PV = nRT \), where \( P \) represents pressure, \( V \) represents volume, \( n \) is the amount of gas in moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

In this exercise, since the helium gas is transferred to a larger container without changing the temperature, the volume increases – potentially altering the dynamics of pressure and molecular movement, which are all interconnected by these laws. Understanding gas laws allows us to predict how gases behave under changing conditions with accuracy.
Disorder and Randomness
Entropy is intricately linked to the concepts of disorder and randomness. It represents the degree of chaos within a system. Imagine you have a neatly stacked pile of cards. If you shuffle them, you've increased the disorder of the cards – in essence, you’ve increased their entropy.

When applying this understanding to gases, if a gas occupies a larger space, its molecules have more room to move around freely and in a random manner. This increased freedom signifies that the molecules are more "disordered" compared to when they were cramped in a smaller space.

This disorder is what increases the entropy of the system when the volume of the gas container is increased, as the molecules can now occupy a greater number of positions. Thus, transferring helium to a larger flask increases its entropy due to the greater possible configurations and randomness of the particles in the larger volume.
Temperature and Volume Relationship
While temperature remains constant in this scenario, it's important to note how temperature generally affects gas volume – a key part of the study of thermodynamics and gas behavior.

According to Charles’s Law, the volume of a gas is directly proportional to its temperature when the pressure is kept constant. This means if you increase the temperature of a gas, its volume also increases if the pressure doesn't change.

However, in the presented exercise, since the temperature remains constant, it does not directly influence the volume increase caused by evacuating the helium into the larger flask. This allows us to focus on the concept of expanding volume as a means to understand entropy, free from other changing variables like temperature.

Often, understanding the relationship between temperature and volume can shed light on how gases will respond in dynamic situations, helping reinforce concepts like why and how entropy increases with volume.

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Most popular questions from this chapter

Which of the following pairs of elements is most likely to create an interstitial alloy? (A) Titanium and copper (B) Aluminum and lead (C) Silver and tin (D) Magnesium and calcium

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

Which of the substances would be soluble in water? (A) Ethylene glycol only, because it has the longest bond lengths (B) Acetone only, because it is the most symmetrical (C) Ethanol and ethylene glycol only, because of their hydroxyl (-OH) (D) All three substances would be soluble in water due to their permanent dipoles.

Directions: Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. Hyprobromous acid, HBrO, is a weak monoprotic acid with a \(K_{\mathrm{a}}\) value of \(2.0 \times 10^{-9} \mathrm{at} 25^{\circ} \mathrm{C} .\) (a) Write out the equilibrium reaction of hyprobromous acid with water, identifying any conjugate acid/based pairs present. (b) (i) What would be the percent dissociation of a 0.50 M solution of hyprobromous acid? (ii) If the 0.50 M solution were diluted, what would happen to the percent dissociation of the HBrO? Why?

The first ionization energy for a neutral atom of chlorine is 1.25 \(\mathrm{MJ} / \mathrm{mol}\) and the first ionization energy for a neutral atom of argon is 1.52 \(\mathrm{MJ} / \mathrm{mol}\) How would the first ionization energy value for a neutral atom of potassium compare to those values? (A) It would be greater than both because potassium carries a greater nuclear charge then either chlorine or argon. (B) It would be greater than both because the size of a potassium atom is smaller than an atom of either chlorine or argon. (C) It would be less than both because there are more electrons in potassium, meaning they repel each other more effectively and less energy is needed to remove one. (D) It would be less than both because a valence electron of potassium is farther from the nucleus than one of either chlorine or argon.

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